MCQ
$\int\limits_1^e {\left( {{x^x} + \log {x^{{x^x}}}} \right)} \,dx = \ ........$
  • A
    $\frac{{e - 1}}{2}$
  • ${e^e} - 1$
  • C
    ${e^e} + 1$
  • D
    ${e^e}$

Answer

Correct option: B.
${e^e} - 1$
$\int^{e}_{1} (x^x+\log x^{x^x})dx$
$ = \int^{e}_{1} (x^x+x^X \log x) dx$
$ = \int^{e}_{1} x^x (1+\log x) dx$
$ x^x = t$ લેતા
$x^x (\log x+1) dx=dt$
$ x=1. t=1$
$x=e, t=e^e$
$ = \int^{e^e}_{1}dx$
$ = [x]^{e^e}$
$ = [e^e-1]$

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